The problem asks to find the general term for the sequence 1, 5, 14, 30, 55, ...

1. Problem Description

The problem asks to find the general term for the sequence 1, 5, 14, 30, 55, ...

2. Solution Steps

Let's analyze the sequence of differences between consecutive terms:

5 - 1 = 4

14 - 5 = 9

30 - 14 = 16

55 - 30 = 25

The sequence of first differences is 4, 9, 16, 25, ... which are squares: $2^2, 3^2, 4^2, 5^2, ...$. This suggests that the original sequence is a cubic polynomial.

Let $a_n$ represent the $n$-th term of the sequence. We assume that $a_n = An^3 + Bn^2 + Cn + D$.

Using the given terms:

For n = 1, $a_1 = A(1)^3 + B(1)^2 + C(1) + D = A + B + C + D = 1$

For n = 2, $a_2 = A(2)^3 + B(2)^2 + C(2) + D = 8A + 4B + 2C + D = 5$

For n = 3, $a_3 = A(3)^3 + B(3)^2 + C(3) + D = 27A + 9B + 3C + D = 14$

For n = 4, $a_4 = A(4)^3 + B(4)^2 + C(4) + D = 64A + 16B + 4C + D = 30$

Subtracting the first equation from the second, the second from the third, and the third from the fourth gives:

$7A + 3B + C = 4$

$19A + 5B + C = 9$

$37A + 7B + C = 16$

Subtracting the first of these new equations from the second, and the second from the third gives:

$12A + 2B = 5$

$18A + 2B = 7$

Subtracting these two equations gives:

$6A = 2$

$A = \frac{1}{3}$

Substituting A into $12A + 2B = 5$ gives:

$12(\frac{1}{3}) + 2B = 5$

$4 + 2B = 5$

$2B = 1$

$B = \frac{1}{2}$

Substituting A and B into $7A + 3B + C = 4$ gives:

$7(\frac{1}{3}) + 3(\frac{1}{2}) + C = 4$

$\frac{7}{3} + \frac{3}{2} + C = 4$

$\frac{14}{6} + \frac{9}{6} + C = \frac{24}{6}$

$\frac{23}{6} + C = \frac{24}{6}$

$C = \frac{1}{6}$

Substituting A, B, and C into $A + B + C + D = 1$ gives:

$\frac{1}{3} + \frac{1}{2} + \frac{1}{6} + D = 1$

$\frac{2}{6} + \frac{3}{6} + \frac{1}{6} + D = 1$

$\frac{6}{6} + D = 1$

$1 + D = 1$

$D = 0$

Thus, $a_n = \frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n = \frac{2n^3 + 3n^2 + n}{6} = \frac{n(2n^2 + 3n + 1)}{6} = \frac{n(2n+1)(n+1)}{6} = \frac{n(n+1)(2n+1)}{6}$

We can also write this as $a_n = \sum_{k=1}^{n} k^2$

3. Final Answer

$a_n = \frac{n(n+1)(2n+1)}{6}$